Difference between revisions of "2003 AMC 10A Problems/Problem 14"
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Since you want <math>n</math> to be the largest number possible, you will want <math>d</math> in <math>10d+e</math> to be as large as possible. So <math>d = 7</math>.Then, <math>e</math> cannot be <math>5</math> because <math>10(7)+5 = 75</math> which is not prime. So <math>e = 3</math>.<math>~~~</math> <math>d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533</math>. | Since you want <math>n</math> to be the largest number possible, you will want <math>d</math> in <math>10d+e</math> to be as large as possible. So <math>d = 7</math>.Then, <math>e</math> cannot be <math>5</math> because <math>10(7)+5 = 75</math> which is not prime. So <math>e = 3</math>.<math>~~~</math> <math>d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533</math>. | ||
So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> ~ MathGenius_ | So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> ~ MathGenius_ | ||
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+ | == Solution 2 == | ||
+ | Since <math>d</math> is a single digit prime number, the set of possible values of <math>d</math> is <math>\{2,3,5,7\}</math>. | ||
+ | |||
+ | Since <math>e</math> is a single digit prime number and is the units digit of the prime number <math>10d+e</math>, the set of possible values of <math>e</math> is <math>\{3,7\}</math>. | ||
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+ | Using these values for <math>d</math> and <math>e</math>, the set of possible values of <math>10d+e</math> is <math>\{23,27,33,37,53,57,73,77\}</math> | ||
+ | |||
+ | Out of this set, the prime values are <math>\{23,37,53,73\}</math> | ||
+ | |||
+ | Therefore the possible values of <math>n</math> are: | ||
+ | |||
+ | <math>2\cdot3\cdot23=138</math> | ||
+ | |||
+ | <math>3\cdot7\cdot37=777</math> | ||
+ | |||
+ | <math>5\cdot3\cdot53=795</math> | ||
+ | |||
+ | <math>7\cdot3\cdot73=1533</math> | ||
+ | |||
+ | The largest possible value of <math>n</math> is <math>1533</math>. | ||
+ | |||
+ | So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> | ||
== See Also == | == See Also == |
Revision as of 11:44, 18 March 2020
Contents
[hide]Problem
Let be the largest integer that is the product of exactly 3 distinct prime numbers , , and , where and are single digits. What is the sum of the digits of ?
Solution 1
Since you want to be the largest number possible, you will want in to be as large as possible. So .Then, cannot be because which is not prime. So . . So, the sum of the digits of is ~ MathGenius_
Solution 2
Since is a single digit prime number, the set of possible values of is .
Since is a single digit prime number and is the units digit of the prime number , the set of possible values of is .
Using these values for and , the set of possible values of is
Out of this set, the prime values are
Therefore the possible values of are:
The largest possible value of is .
So, the sum of the digits of is
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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